Question 882289
<pre>
Here is a way that shows even more detail:

{{{(5/(2x^2-18) - 2/(x^2-7x+12))/(2/(x^2-x-12) - 4/(3x^2-27))}}}

As the good doctor above says, factor all the four denominators

Factor: {{{2x^2-18=2(x^2-9)=2(x-3)(x+3)}}}
Factor: {{{x^2-7x+12=(x-3)(x-4)}}}
Factor: {{{x^2-x-12=(x+3)(x-4)}}}
Factor: {{{3x^2-27=3(x^2-9)=3(x-3)(x+3)}}}

{{{(5/(2(x-3)(x+3)) - 2/((x-3)(x-4)))/(2/((x+3)(x-4)) - 4/(3(x-3)(x+3)))}}}

The LCD of all those denominators is {{{red(6(x-3)(x+3)(x-4))}}}

Put it over 1: {{{expr(red((6(x-3)(x+3)(x-4))/1))}}}

and multiply every term in the numerator and denominator by it:

{{{(expr(5/(2(x-3)(x+3)))expr(red((6(x-3)(x+3)(x-4))/1)) - expr(2/((x-3)(x-4)))expr(red((6(x-3)(x+3)(x-4))/1)))/(expr(2/((x+3)(x-4)))expr(red((6(x-3)(x+3)(x-4))/1)) - expr(4/(3(x-3)(x+3)))expr(red((6(x-3)(x+3)(x-4))/1)))}}}

Then you cancel as much as you can:

{{{(expr(5/(cross(2)(cross(x-3))(cross(x+3))))expr(red((""^3cross(6)(cross(x-3))(cross(x+3))(x-4))/1)) - expr(2/((cross(x-3))(cross(x-4))))expr(red((6(cross(x-3))(x+3)(cross(x-4)))/1)))/(expr(2/((cross(x+3))(cross(x-4))))expr(red((6(x-3)(cross(x+3))(cross(x-4)))/1)) - expr(4/(cross(3)(cross(x-3))(cross(x+3))))expr(red((""^2cross(6)(cross(x-3))(cross(x+3))(x-4))/1)))}}}

{{{(15(x-4)-12(x+3))/(12(x-3)-8(x-4))}}}

{{{(15x-60-12x-36)/(12x-36-8x+32)}}}

{{{(3x-96)/(4x-4)}}}

You can leave it like that or else you can factor out a common
factor in the numerator and denominator:

{{{(3(x-32))/(4(x-1))}}}

Edwin</pre>